# Is a mixed second rank tensor reducible?

As a complete novice, I'm reading a text which says that a mixed second rank tensor $$T^{u}_{v}$$ is reducible but don't see how. Anyone care to show me? ## Answers and Replies

robphy
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What is the definition of reducible? (I'm not sure what you mean.)

What is the definition of reducible? (I'm not sure what you mean.)

It means it can be broken down into parts that transform among themselves.

robphy
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Can you provide the reference where the statement in your first post appears?

Can you provide the reference where the statement in your first post appears?

It comes from Anderson's Principles of Relativty Physics on page 19:

" Thus from the components of a tensor T$$^{uv}$$ we can construct its symmetric part T$$^{(uv)}$$ and its antisymmetric part T$$^{[uv]}$$ according to

$$T^{(uv)} = 1/2(T^{uv}+ T^{vu})$$

$$T^{[uv]} = 1/2(T^{uv}- T^{vu})$$

Similarly we can construct the transformed symmetric part T$$^{'(uv)}$$ and the antisymmetric part T$$^{'[uv]}$$ from the transformed T$$^{'uv}$$. Then, one can show that T$$^{'(uv)}$$ is a function of T$$^{(uv)}$$ and the mapping function only, and similarly for T$$^{[uv]}$$. When ever a geometrical object can be broken up into parts that transform among themselves, we say that we have a reducible object. If no such decomposition as possible. we have an irreducible object.

robphy
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OK, I see now.
I take it you are referring to Problem 1.7....
where they are telling you to "decompose" $T_\nu{}^\mu$
into irreducible parts consisting of
its "trace" $T_\mu{}^\mu$ and
a "traceless " (or "trace-free") tensor (which they give as $T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho$ on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write $T_\nu{}^\mu$
as a sum of two tensors, one of which they gave you:
$T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu$

OK, I see now.
I take it you are referring to Problem 1.7....
where they are telling you to "decompose" $T_\nu{}^\mu$
into irreducible parts consisting of
its "trace" $T_\mu{}^\mu$ and
a "traceless " (or "trace-free") tensor (which they give as $T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho$ on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write $T_\nu{}^\mu$
as a sum of two tensors, one of which they gave you:
$T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu$

Yes. The question says: "show that it is reducible *and* its irreducuble parts are...". So I would have thought you should be able to first show its reducible from the transformation properties of a mixed second rank tensor, rather than being told what the irreducible components are and from that showing its reducible :smile